The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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Equivalence of countable choice for subsets of the reals and “second countable $\implies$ Lindelöf”

Looking for a proof that in a second countable space every open cover has a countable subcover -- i.e. every s.c. space is a Lindelöf space -- I bumped into this question. That answered my question. I ...
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Prove there exists a strictly increasing function from the natural numbers to a partially strictly ordered set

Let $P$ be a non-empty partially strictly ordered set and assume no element of $P$ is maximal (i.e. for every $x \in P$ there exists $y\in P$ with $x < y$). Show there exists a function ...
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Every II-finite set is III-finite

I need some help proving that if a set $X$ is II-finite then it is III-finite, i.e. if every non-empty family of subsets of $X$ which is linearly ordered by inclusion has a maximal element under ...
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Lebesgue measure without choice

From this question and this question (and their answers) I gather that it is consistent with ZF without The Axiom of Choice to assume that there exist countable sets $A_n$, $n\in \mathbb N$, such that ...
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Accumulation point is a limit of some sequences in first-countable space with Axiom of Choice.

I have a statement which says that Let $X$ be a first-countable space and $E$ be a subset of $X$. If $p$ is an accumulation point of $E$, then there exists a sequence in $E$ that converges to ...
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1answer
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Finite complement topology over $\Bbb{R}$ is not second-countable under ZF?

Under $\mathsf{ZF+AC_\omega}$, the space $\Bbb{R}$ with finite complement topology is not second-countable, since a countable union of countable subsets of $\Bbb{R}$. However, such argument doesn't ...
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Breaking the AC barrier using Kelley-Morse set theory

In her blog post Variants of Kelley-Morse set theory, Prof. Gitman proves that every model of the the common version of Kelley-Morse set theory ($\mathsf{KM}$) is a model of the Wikipedia version of ...
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Every set with more than point admits a permutation with no fixed point and the Axiom of Choice [duplicate]

Assuming axiom of choice , for any set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ . Is the converse true , i.e. Does the statement " ...
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Consequences of the negation of the Axiom of Dependent Choice

It seems to me that a proper reason to include The Axiom of Choice as a foundational axiom of set theory should be based on the observation that the negation of The Axiom of Choice has absurd ...
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$\forall \alpha \exists \beta: \beta > \alpha$ where $\alpha$ and $\beta$ cardinals

I have to prove ZF $\vdash$ $\forall \alpha \exists \beta:\beta > \alpha$, where $\alpha, \beta-$ cardinal numbers. I can prove it only in ZFC. Let's fix some cardinal number $\alpha$. By ...
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In every ring with unity satisfying ACC, every ideal is finitely generated; can we prove it without assuming Axiom of choice?

Assuming Zorn's lemma, we can prove that in every ring with unity satisfying Ascending-Chain-Condition, every ideal is finitely generated. Is this statement equivalent to Zorn's lemma? Can we prove it ...
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In any commutative ring with unity, every prime ideal is finitely generated implies every ideal is finitely generated; can it be prove without A.C.?

Assuming Zorn's lemma, "In any commutative ring with unity, if every prime ideal is finitely generated, then every ideal is finitely generated". Is the converse true, i.e. if in any commutative ring ...
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1answer
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Assuming the axiom of choice ,how to prove that every uncountable abelian group must have an uncountable proper subgroup?

Assuming the axiom of choice , how to prove that every uncountable abelian group must have an uncountable proper subgroup ? Related to Does there exist any uncountable group , every proper subgroup ...
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1answer
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Can you verify this proof of the Schroeder-Bernstein theorem?

I'm a freshman in college and my professor challenged us to find a proof of this theorem. Please don't give me the answer but please verify if this proof works or, if not, if it is the start of a ...
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1answer
62 views

Proving that “Every non-trivial ring (i.e. with more than one element ) with unity has a maximal ideal” implies axiom of choice is true

I know that assuming axiom of choice or equivalently Zorn's lemma , it can be proved that every non-trivial ring with unity has a maximal ideal (two sided ) . The wiki article on axiom of choice says ...
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1answer
30 views

Formal proof that $N$ is the smallest infinite set [duplicate]

I wish to write a formal proof of the following statement: For any infinite set $X$, there exists an injection $f:\mathbb{N}\to X$. I'd like the proof to explicitly use the full axiom of choice (for ...
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1answer
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Acyclicity of Flasque sheaves without A.C.

I say that a sheaf on a space X, is flasque if the restriction maps are surjective, that is any local section extend to a global section. Now it is a fact that if $F_1$ is flasque and if $0 \to F_1 ...
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Cardinality of sets of reals without choice

Assuming just ZF (no axiom of choice): Does $\aleph_n\leq|\mathbb{R}|$ for all $n<\omega$ imply $\aleph_\omega\leq|\mathbb{R}|$? (with $\kappa\leq|\mathbb{R}|$ meaning that there is a set of reals ...
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1answer
25 views

Conceptual problem in Noetherian rings and Zorn Lemma

I've got a problem, and it's that I can't see the difference of one of the definitions of Noetherian ring, and supposing the zorn lemma for that ring. I mean, if I use the definition of the maximal ...
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130 views

Closure of a set in a “Topology of finite complement”

Well, I was reading this article by Kelley and when reached the point where he say that $X_a$ is closed in $Y_a$ I had to stop, probably mine is just a stupid misunderstand but can't figure out how to ...
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Did I use axiom of choice in my proof?

I have two different affine open covers for a scheme $X$, say $X = \cup_{i \in I} U_i$ and $X = \cup_{j \in J} V_j$. For each $p \in X$, we know there exist some $i(p)$ and $j(p)$ such that $p \in ...
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On axiom of choice [on hold]

Suppose we have a countable set of intervals (i.e. I= $\{[a,b]|a,b \in \mathbb{R}\}$, does axiom of choice guaranteed us to find one with the maximal length?
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Every epimorphism in Sets is split: why is it equivalent to axiom of choice?

Suppose that $f: A \rightarrow B$ is epic in Sets. One can construct a section $s: B \rightarrow A$ of $f$ as follow: Let us define an equivalence relation $R$ on $A$ as follow: $aRa'$ iff $a, a' \in ...
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1answer
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Forcing reference

Who first proved that, over ZF, the statement (1) The reals are well-orderable is strictly stronger than the statement (2) Every real-indexed family of nonempty sets of reals has a choice ...
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1answer
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Every Partially ordered set has a maximal independent subset

I am working on this problem:- Every Partially ordered set has a maximal independent subset. Definition:Let $\langle E,\prec\rangle$ be a partially ordered set. A subset $A\subset E$ is called ...
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Why in Teichmüller-Tukey lemma finitness is essential?

First we will state a Teichmüller-Tukey Lemma: Let $A$ be a set and $\phi$ be a property defined on all finite subset of $A$. Assume that $B$ is a subset of $A$ such that each finite subset of $B$ ...
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1answer
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Finitely-additive measure over $\Bbb{N}$

On the set of natural number, we can consider the finitely-additive measure defined as: $$\mu(A) = \lim_{n\to\infty}\frac{\#(A\cap [1,n])}{n}.$$ However, there is a definable (by PA, or some ...
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Is the axiom of choice used here?

I was reading the following How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles? The answer by "user4594" basically considers a k-cell with rational ...
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One more AC equivalence question

Is "Every vector space admits a norm" weaker than AC? I know that the statement follows from "Every vector space has a basis", which is equivalent to AC.
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Example of a maximal idea

Let $A$ be the set of bounded continuous functions from the set of real numbers to itself. Then $A$ is a ring under pointwise addition and multiplication. The set $I$ of all functions $f \in A$ ...
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1answer
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Question related to ordinal number without using Axiom of Choice.

Can we proof this result without using Axiom of Choice :- $$A\cap \alpha=\emptyset \,\,\,\, \mbox{and}\, \, \, A\times \alpha \sim A\cup \alpha$$ then there is an $A^{'} \subset A$ such that $\alpha ...
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Is this an accurate way to write the axiom of choice using transversals?

Is this an accurate way to write the axiom of choice using transversals? $\forall \mathcal{C} \exists T\, (T\subseteq \cup\,\mathcal{C} \land \forall X \exists s \forall t\, ((X\in\mathcal{C} \land ...
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Outer automorphisms of the infinite symmetric group

Denote by S$_\infty$ the group of permutations of $\mathbb N$. Question: Does there exist an outer automorphism of S$_\infty$, and if so, can one be exhibited? Does this depend on the continuum ...
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$Gal(\bar{\mathbb Q}/\mathbb Q)$ without choice, and constructive Galois theory

By this question: Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice? We have that over ZF, algebraic closures of $\mathbb Q$ aren't unique. Are their Galois groups as extensions over ...
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Recursive use of the Axiom of Choice

In a standard proof that any sequence-compact metric space $(X,d)$ has a (finite) $\varepsilon$-net, the approach is the following: Make a sequence $(x_n)$ such that $$ x_{n+1}\notin\bigcup_{i=1}^n ...
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What is the name of proofs with (without) Axiom of Choice

In many contexts we distinguish between proofs using AC and proofs which do not use AC. (To phrase this somewhat differently: If there is a proof without AC, this proof is usually preferred.) I would ...
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Are countably infinite, compact, Hausdorff spaces necessarily second countable?

Let $X$ be a compact Hausdorff space. If $X$ is finite, then there are at most finitely many open sets and so it is second-countable. On the flipside if $X$ is uncountably infinite, it is possible ...
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The topological product of path-connected spaces is path-connected $\Rightarrow\sf AC$?

There is a very natural proof that the product of path-connected spaces is path-connected: Let $X=\prod_{i\in I}X_i$ be a product of path-connected spaces $X_i$. Given $(x_i)_{i\in I},(y_i)_{i\in ...
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Proving that axiom of choice is equivalent to this proposition

I'm trying to do the problem 8.2 of the book Notes on Set Theory of Moschovakis. I want to prove that Axiom of Choice is equivalent to this statement: $$\forall A\neq \emptyset,\, \forall ...
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A descending chain and choice axiom.

Let $R$ be a relation on a set $A$ such that for every $x\in A$ there exist a element $y\in A$ such that $x\mathrel{R}y$ the there exist a function $f\colon\omega\rightarrow A$ such that ...
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1answer
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Is this proof using the axiom of choice correct?

In Munkres' Topology, the axiom of choice is defined as follows: Given a collection $A$ of disjoint nonempty sets, there exists a set $C$ containing exactly element from each element of $A$; that ...
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Choice-less Set Theory for Dummies

In almost every graduate set theory text there are some parts about equivalences of $AC$, its consequences, some axioms like $AD$ which imply $\neg AC$, some well-known axiomatic systems which $AC$ ...
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Construction of injective hulls without axiom of choice

Motivation: It is known that without the axiom of choice (AC), it is not provable that all categories of modules have enough injectives, let alone injective hulls. Still, there are examples of rings ...
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Do I need axiom of choice in this question?

If $X$ is a finite set and $Y$ an infinite one, I'm trying to prove there is a injective function $f:X\to Y$. My solution: Let $X=\{x_1,\ldots,x_n\}$, for each $x_i$ choose $y_1,\ldots,y_n$ such that ...
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Why is Zorn's Lemma called a lemma?

By convention a Lemma is a technical intermediate step which has no standing as an independent result. Lemmas are only used to chop big proofs into handy pieces. (Quoted from here) I am wondering why ...
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if $f:X\to Y$ is a surjective, then there is an injective function $g:Y\to X$ [duplicate]

I'm trying to prove if $f:X\to Y$ is a surjective, then there is an injective function $g:Y\to X$. I know this intuitively, since $f$ is surjective, for every $y$ in $Y$ we need just to choice an ...
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1answer
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Proving statements , that require Zorn's lemma , for countable case directly by well-ordering principle of natural numbers

We know that for countable sets , the existence of a choice function is a consequence of the well-ordering principle ; and it is also known that the results like "every vector space has a maximal ...
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How do I prove that $S_A\cong S_B\implies |A|=|B|$?

Let $A,B$ be infinite sets such that $S_A\cong S_B$. (Symmetric groups are group isomorphic) How do I prove that $|A|=|B|$? The only proof I know uses Axiom of choice. (That is, using AC to give ...
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1answer
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Additive function and continuity at a point

Does continuity at a point and Additive function imply continuity at all other points in a normed linear space. Is there some result like there exist a in field such that f(x) = ax for all x in normed ...
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Basic definition of continuity [duplicate]

Ltf(c+h) = f(c)(h goes to 0) if and only if Ltf(x) = f(c)(x goes to c). I am able to prove this fact using sequential criterion of continuity. But sequential criterion is dependent on Axiom of ...